I'm posing this problem partly because I think some folks on this site would be interested in working on it and partly because I would like to see a purely combinatorial proof. (But please post any proofs; I would be interested in noncombinatorial ones, too. I've learned a lot on this site by reading alternative proofs of results I already know.)

I'll wait a few days to give others a chance to respond before posting my proofs.

What immediately jumps out is how such a complicated function like the totient function has an identity involving the simple summation of integers 1 to n. Seems like a good way to prove it is to show that the expression inside the sum is a "rearrangement" of the sequence of integers 1 to n
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crasicOct 27 '10 at 5:42

The sum of elements in column $j$ is $\displaystyle \left\lfloor \frac{n}{j} \right\rfloor \varphi(j)$ and the identity just says the total sum by summing the rows is same as the total sum by summing the columns.

Both sides count the number of fractions (reducible or irreducible) in the interval (0,1] with denominator $n$ or smaller.

For the right side, the number of ways to pick a numerator and a denominator is the number of ways to choose two numbers with replacement from the set $\{1, 2, \ldots, n\}$. This is known to be
$$\binom{n+2-1}{2} = \frac{n(n+1)}{2}.$$

Now for the left side. The number of irreducible fractions in $(0,1]$ with denominator $k$ is equal to the number of positive integers less than or equal to $k$ and relatively prime to $k$; i.e., $\varphi(k)$. Then, for a given irreducible fraction $\frac{a}{k}$, there are $\left\lfloor \frac{n}{k} \right\rfloor$ total fractions with denominators $n$ or smaller in its equivalence class. (For example, if $n = 20$ and $\frac{a}{k} = \frac{1}{6}$, then the fractions $\frac{1}{6}, \frac{2}{12}$, and $\frac{3}{18}$ are those in its equivalence class.) Thus the sum
$$\sum_{k=1}^n \left\lfloor\frac{n}{k} \right\rfloor \varphi (k)$$
also gives the desired quantity.

Look at the problem of counting all the triples (d,m+1,k) where $d\leq m\leq n$ and $gcd(d,m)=\frac{m}{k}$.
For each (d,m+1) we have exactly one k such that $gcd(d,m)=\frac{m}{k}$, and so we are counting exactly all the pairs (i,j) where $ i < j \leq n+1 $ which is $\binom{n+1}{2}$.

On the other hand, for each k , if $ gcd(d,m)=\frac{m}{k} $, then $ k \mid m $ and there are $ \left\lfloor \frac{n}{k}\right\rfloor $ such m's, and for each m we have $\varphi(k)$ d's that satisfies this equality, so there are $ \left\lfloor \frac{n}{k}\right\rfloor \varphi(k) $ triples that end with k.

The idea is counting the number of ways to choose two numbers from 1,2,...,n+1 (which is the right side of the equation). Once you choose the larger number m+1, you have m options which is exactly $\sum_{d\mid m} \varphi(d)$ (this is basically what Moron wrote).
What I wanted to do is to change the counting according to the "relation" between d and m, and the one that worked out is that $\frac{m}{gcd(d,m)}=k$.

So, going over triples (d,m+1,k) where $\frac{m}{gcd(d,m)}=k$ and $d\leq m\leq n$ just mean that (d,m+1) is an ordered pair from {1,...,n} and k is decided as above.

On the other hand, for specific k, if $\frac{m}{gcd(d,m)}=k$ then $k \mid m$ and there are $ \left\lfloor \frac{n}{k}\right\rfloor$ such m's. For each m, if $gcd(d,m)=m/k$ then $gcd(\frac{d}{m/k},\frac{m}{m/k})=gcd(\frac{d}{m}k,k)=1$ and there are $\varphi(k)$ such d's (because $d\leq m$).

So, or each k there are $ \left\lfloor \frac{n}{k}\right\rfloor \varphi(k) $ options , and summing over k will give the left side of the equation.